The `pH` of an aqueous solution of `Ba(OH)_(2)` is `10`. If the `K_(sp)` of `Ba(OH)_(2)` is `1xx10^(-9)`, then the concentration of `Ba^(2+)` ions in the solution in `mol L^(-1)` is

Options

A. `1xx10^(-2)`
B. `1xx10^(-4)`
C. `1xx10^(-1)`
D. `1xx10^(-5)`

Answer

C

Explanation

`pOH =14-10=4`
`:. [OH^(-)] =1xx10^(-4) M`
`K_(sp)=[Ba^(2+)][OH^(-)]^(2)`
or `[Ba^(2+)]=K_(sp)//[OH^(-)]^(2)=(1xx10^(-9))/((1xx10^(-4))^(2))`